# Does something weigh less 1km under the ground?

1. Jul 24, 2010

### nonequilibrium

Okay, the title sounds weird, let me explain:

The father of a friend of mine used to work in the mines about 800m under the ground and he always experienced that when he had to carry heavy gears, it felt as if the item was considerably lighter to carry underground than when he was aboveground carrying the same item.

Is there something that could explain this?

2. Jul 24, 2010

### D H

Staff Emeritus
Lighter? Yes. Considerably lighter? No. The effect is *tiny*. Let's say your father's friend had to carry 50 pounds of equipment. Going 1/2 mile (~800m) underground would make that equipment feel about 1/10 of an ounce lighter.

3. Jul 24, 2010

### Dickfore

Inside a homogeneous sphere, using Gauss' Law, one can show that the gravitational field changes as:

$$g(r) = \frac{g_{0} \, r}{R}$$

where $g_{0}$ is the gravitational field on the surface of the sphere R is the radius of the sphere and r is the distance from the center. Taking:

$$r = R - h, \; h \ll R$$

then the relative change in the gravitational field is:

$$\delta = \frac{\Delta g}{g} = \frac{h}{R}$$

Because $R = 6.37 \times 10^{6} \, \mathrm{m}$, it means that $\delta = 1.3 \times 10^{-4}$.

I highly doubt that any human is sensitive enough to such small variations in the local gravitational field.

Last edited: Jul 24, 2010
4. Jul 24, 2010

### K^2

Actually, because the crust is less dense than the core, for the first kilometer, things keep getting heavier. I don't remember exact depth past which the things start getting lighter, but it is significantly deeper than that.

The effect might be due to higher atmospheric pressure allowing for more oxygen in the blood, making things feel lighter. At 800m bellow the surface, pressure is going to be 10% higher than at the surface. Unlike changes in gravity, that is significant.

5. Jul 24, 2010

### nonequilibrium

Yeah thank you guys, so there's no real physical way to explain his sensations.

K^2 might be on the right track with his biological explanation -- too bad I hardly know enough biology to say how significant the change in apparent strength is due to oxygen increase. Should I ask the same question in the Biology forum?

6. Jul 24, 2010

### K^2

You can try. I know that reduced pressure/oxygen makes you feel fatigued. But I don't know how significant the reverse is going to be.

7. Jul 24, 2010

### pallidin

Not so.
As you move down into the earth, there is more mass above you, pulling you slightly upwards. Therefore, you would weigh less.

8. Jul 24, 2010

### K^2

Please tell me that words "Gauss Theorem" make the light bulb light up. I don't want to type it all out.

9. Jul 24, 2010

### pallidin

Huh?

10. Jul 24, 2010

### nonequilibrium

pallidin, it can be proven (rather elegantly that in a hollow sphere there's no gravitational attraction because the vectors cancel out exactly for every point inside the sphere

11. Jul 24, 2010

### pallidin

Of course there is gravitational attraction!
Just because force vectors "cancel", this DOES NOT mean that the force no longer exists. In fact, for gravity, you can not shield it at all.
The "effect" is what is moderated, not gravity itself.

12. Jul 24, 2010

### nonequilibrium

I get your point and it's a fair one; I was just using "no gravitational attraction" as shorthand for "no net gravitational attraction".

But anyway, come to think of it, in the theory of Einstein, it seems like there really is no gravitational attraction, but maybe I shouldn't say anything about that because I'm not even sure. But anyway, the theorem of Gauss says the effect of gravity is zero inside a hollow sphere, demolishing your earlier argument that you're pulled up underneath the earth's crust.

13. Jul 24, 2010

### K^2

pallidin, each spherical shell produces a gravitational field identical to point mass with the combined mass of the shell for any object outside the shell, and precisely zero anywhere in side the shell. Do you need that proven? Or can you look up Gauss Theorem on your own?

Now, given some density depending only on r, ρ(r), the equation for acceleration due to gravity at distance r from center of the planet can be given by the following.

$$g(r)=\frac{G}{r^2}\int_{0}^{r}\rho(z)dz$$

Is that equation clear? Do you need me to show plots to demonstrate that it keeps growing after you started going underground, if ρ is higher at the core then at the surface?

14. Jul 24, 2010

### pallidin

I give up. There is no such thing as zero gravitational field regarding a hollow, massive object. Vectors could cancel? Yes, with regard to effect.
Does the field cease to exist? No.

15. Jul 24, 2010

### K^2

It can't and screw 200 years of physics and mathematics that say otherwise?

16. Jul 24, 2010

### nonequilibrium

pallidin, you're being stubborn to no cause: you're the one going into the debate of there actually being any force or not. That is not what we're saying or care to say: all we're saying is that the vectors cancel, and that is all that counts in this discussion. You were saying a person 1km below ground would experience a net upward force. This is false, the person experiences nothing what is above him.

17. Jul 24, 2010

### pallidin

With all due respect, that simply is not true at all.

18. Jul 24, 2010

### nonequilibrium

I don't think you understand the concept of vector. Do you understand the vector sum of gravitational force is zero everywhere inside? If you accept that, there follows immediately that you won't experience any force, because what is experiencing a force? Accelerating. What is accelerating? Have a non-zero force vector act on you.

And if you don't accept the fact that the vector sum is zero, well, that's simply Gauss' theorem and that's been proven.

I don't know what else to say?

19. Jul 24, 2010

### K^2

And you are wrong. Have you looked up Gauss Theorem yet? Do so. Really.

20. Jul 24, 2010

### pallidin

Ok, you are defending a wrong perspective.
I'll do this gently...

Imagine that you are a water-filled balloon in the center of a hollow earth.
The water IN THAT BALLOON WILL BE PULLED away(slightly) from it's center and optimize on the spherical edge inside the envelope of the balloon.
Get it?